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Commit ece0ac4

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Sync LeetCode submission Runtime - 94 ms (14.45%), Memory - 36.3 MB (58.17%)
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<p>You are given an integer <code>eventTime</code> denoting the duration of an event, where the event occurs from time <code>t = 0</code> to time <code>t = eventTime</code>.</p>
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<p>You are also given two integer arrays <code>startTime</code> and <code>endTime</code>, each of length <code>n</code>. These represent the start and end time of <code>n</code> <strong>non-overlapping</strong> meetings, where the <code>i<sup>th</sup></code> meeting occurs during the time <code>[startTime[i], endTime[i]]</code>.</p>
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<p>You can reschedule <strong>at most</strong> <code>k</code> meetings by moving their start time while maintaining the <strong>same duration</strong>, to <strong>maximize</strong> the <strong>longest</strong> <em>continuous period of free time</em> during the event.</p>
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<p>The <strong>relative</strong> order of all the meetings should stay the<em> same</em> and they should remain non-overlapping.</p>
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<p>Return the <strong>maximum</strong> amount of free time possible after rearranging the meetings.</p>
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<p><strong>Note</strong> that the meetings can <strong>not</strong> be rescheduled to a time outside the event.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">eventTime = 5, k = 1, startTime = [1,3], endTime = [2,5]</span></p>
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<p><strong>Output:</strong> <span class="example-io">2</span></p>
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<p><strong>Explanation:</strong></p>
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<p><img alt="" src="https://assets.leetcode.com/uploads/2024/12/21/example0_rescheduled.png" style="width: 375px; height: 123px;" /></p>
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<p>Reschedule the meeting at <code>[1, 2]</code> to <code>[2, 3]</code>, leaving no meetings during the time <code>[0, 2]</code>.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">eventTime = 10, k = 1, startTime = [0,2,9], endTime = [1,4,10]</span></p>
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<p><strong>Output:</strong> <span class="example-io">6</span></p>
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<p><strong>Explanation:</strong></p>
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<p><img alt="" src="https://assets.leetcode.com/uploads/2024/12/21/example1_rescheduled.png" style="width: 375px; height: 125px;" /></p>
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<p>Reschedule the meeting at <code>[2, 4]</code> to <code>[1, 3]</code>, leaving no meetings during the time <code>[3, 9]</code>.</p>
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</div>
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<p><strong class="example">Example 3:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">eventTime = 5, k = 2, startTime = [0,1,2,3,4], endTime = [1,2,3,4,5]</span></p>
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<p><strong>Output:</strong> <span class="example-io">0</span></p>
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<p><strong>Explanation:</strong></p>
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<p>There is no time during the event not occupied by meetings.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= eventTime &lt;= 10<sup>9</sup></code></li>
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<li><code>n == startTime.length == endTime.length</code></li>
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<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= k &lt;= n</code></li>
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<li><code>0 &lt;= startTime[i] &lt; endTime[i] &lt;= eventTime</code></li>
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<li><code>endTime[i] &lt;= startTime[i + 1]</code> where <code>i</code> lies in the range <code>[0, n - 2]</code>.</li>
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</ul>
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class Solution:
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def maxFreeTime(
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self, eventTime: int, k: int, startTime: List[int], endTime: List[int]
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) -> int:
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n = len(startTime)
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res = 0
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t = 0
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for i in range(n):
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t += endTime[i] - startTime[i]
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left = 0 if i <= k - 1 else endTime[i - k]
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right = eventTime if i == n - 1 else startTime[i + 1]
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res = max(res, right - left - t)
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if i >= k - 1:
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t -= endTime[i - k + 1] - startTime[i - k + 1]
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return res

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